Monday, October 20, 2014

GTO Brainteaser #8 Bonus Solution -- Optimal Betsize Calculations

I got a request for a numeric answer to the bonus of GTO Brainteaser # 8 which involves solving for an optimal bet size in a two street bluffing game. The solution to the non-bonus question is here and is worth reading first. I haven't actually done a post on deriving optimal betsizing in multistreet play before so I thought it would be useful to demonstrate the mathematics involved. I'm not going to restate the game structure again here so please check out the original problem statement if you are not familiar with the original game.

The basic technique for calculating optimal bet sizing is as follows.

Rather than using a fixed betsize in your calculations, make the bet size a variable and solve for GTO strategies as a function of that variable

Compute the EV of the game when both players play the GTO strategies as a function of the bet size variable

Maximize the EV of the person making the bet with respect to the bet size variable. That is your optimal bet size.

While this technique is quite simple conceptually, the actual algebra involved can be hairy so I usually just make wolfram alpha do it. So lets get started.

To calculate the optimal betsize we will make a few assumptions that are reasonably easy to verify and that I have shown in other posts / videos.

The hero should always bet the nuts on the turn. This allows him to "compound the nuts" over multiple streets which as I showed here and in more depth in my CardRunners videos is always +EV compared to betting on a single street with a polarized range.

On the river it will always be most profitable to shove with the nuts with our polarized range. This is quite simple to prove and I showed it in my first CardRunner's video.

The hero's EV with his air on the river will be 0 unless z is so large that it is optimal for the villain to always fold the river. However, clearly if the villain were always folding in a spot where the hero might hold air on the river, he would never call a turn bet, so any time a turn bet is called, the hero's EV with air must be 0 on the river

Combining observations 1 and 2 we can parameterize our bet sizing strategy with a single number x, the number of chips that we plan to bet on the turn.

If we bet x chips on the turn, then we know we will jam the river and bet the rest of our chips. Given a starting turn pot of 100 chips if we bet x chips on the turn, the river pot will be 100 + 2x when we are called.

Our river jam will thus be a bet of (150 - x) into a pot of 100 + 2x. This means that we will be making a

b = (150-x) / (100 + 2x) percentage pot bet on the river.

Now lets call the frequency of the villain calling the turn c. Observation 4 tell us that since the hero's river EV is 0 with air, his EV for bluffing the turn is very simple to calculate.

EV[turn bluff] = (1-c) * 100 - c * x

Clearly the villain always folding or always calling the turn is highly exploitable so we know his turn calling strategy is mixed and we can apply indifference conditions to see that

c = 100 / (100 + x)

What about the villains calling frequency on non 3d/2c rivers? This will of course just be determined by indifference conditions that depend on the pot / bluff size. Call the villains river calling frequency rc.

EV[river bluff] = (1 - rc) * (100 + 2x) - (150 - x) * rc

Indifference conditions imply that

rc = (2x + 100) / (x + 250)

Now we can write the optimal turn bluffing frequency as a function of the turn bet size as well by looking at the villains EV for calling. I calculated the villains EV of calling when the bet size was 50 chips in my previous post but I will duplicate the calculation here, assuming that the hero is bluffing with frequency with his air and always betting the nuts. This means that (1-z) = a / (1 + a) of his betting range is air and z = 1 / (1 + a) of his betting range is the nuts.

Since the hero's EV with air on all rivers is 0, when he bluffs and we call we win the 100 chip pot plus his turn bet size in EV. When he holds the nuts on 3c/2d runouts our EV is 0 on the river and on other runouts our EV when our opponent holds the nuts .

If we apply indifference conditions to say that the EV of a call must be 0, this a relationship between z and x that we can solve for z.

Wolfram Alpha is much better at algebra than me so I just computed that relationship here.

Now the EV of the game for the villain is just how often the hero checks the turn, which is just (1-a) / 2, because by indifference conditions, when the hero bets the villain is indifferent between calling and folding and thus his EV is 0.

Since z = 1 / (1 + a), a = (1/z) - 1, so (1 - a)/2 = (2 - 1/z) / 2

So the EV of the game is (2 - 1/z) / 2 for the villain and we know z as a function of x. Thus the optimal bet size for the hero is the value of x that minimizes his opponents EV, (2 - 1/z) / 2, where z is between 1/2 and 1 (because our betting range is at least 1/2 nuts and at most 100% nuts. Since this is clearly decreasing in z, we just need to minimize z.

Again I calculated this using wolfram alpha here. The result is that the optimal bet size is 52.69 chips. This intuitively makes sense, as we would expect to bet slightly larger on the turn with some of the river runouts killing our action than we would without that risk.

The EV of this game is ~87.89 so the EV gain by changing betsize in this case is tiny, about 0.02 chips.